Answer:
not true because the mass from the heavy car will cause it to damage more
Explanation:
Answer:
T= 4.24sec
Explanation:
We are going to use the formula below to calculate.

Where T is period
L is length of rod
g is acceleration due to gravity =
From the problem, the rod is pivoted at 1/4L which means that three quarter of the rod was used for the oscillation. lets call this

= 4.4625m
thus
T= 4.24sec
Answer:
option "c"
Explanation:
because in gases molecules are further apart and move very quickly
Answer:
The coefficient of static friction is 0.29
Explanation:
Given that,
Radius of the merry-go-round, r = 4.4 m
The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.
We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

v is the speed of cat, 

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.
Answer:
(a) 0.063 m/s
(b) 1.01 m/s
Explanation:
rate of volume flow, V = 4 x 10^-6 m^3/s
(a) radius, r = 4.5 x 10^-3 m
Let the speed of blood is v.
So, V = A x v
where A be the area of crossection of artery
4 x 10^-6 = 3.14 x 4.5 x 10^-3 x 4.5 x 10^-3 x v
v = 0.063 m/s
Thus, the speed of flow of blood is 0.063 m/s .
(b) Now r' = r / 4 = 4.5 /4 x 10^-3 m = 1.125 x 10^-3 m
Let the speed is v'.
So, V = A' x v'
4 x 10^-6 = 3.14 x 1.125 x 10^-3 x 1.125 x 10^-3 x v'
v' = 1.01 m/s
Thus, the speed of flow of blood is 1.01 m/s .